Question

The differential equation of the family of curves $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{a^2} + {\lambda ^2}}} = 1$ ($\lambda $is arbitrary constant)
A.$({x^2} - {a^2}){y_1} = xy$
B.$({x^2} - {a^2}){y_2} - xy = 0$
C.${x^2}{y_2} - {a^2}y = 0$
D.$({x^2} - {a^2}){y_1} + xy = 0$

Answer 1

Hint: The differential equation can be found by eliminating $\lambda $from the given family of curves. As there is only one variable that needs to be eliminated, we will first transform the equation so that the $\lambda $that is to be eliminated is on one side and the variables on others. By differentiating both sides, we will eliminate the variable and find the differential equation in terms of$x,y,{y_1},a$.

Complete step by step solution:

Given curve: $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{a^2} + {\lambda ^2}}} = 1$
We will transform the curve into a simpler equation so that eliminating $\lambda $could be easy.
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{a^2} + {\lambda ^2}}} = 1$
$\dfrac{{{y^2}}}{{{a^2} + {\lambda ^2}}} = 1 - \dfrac{{{x^2}}}{{{a^2}}}$
$\dfrac{{{y^2}}}{{{a^2} + {\lambda ^2}}} = \dfrac{{{a^2} - {x^2}}}{{{a^2}}}$
Cross multiplying,
${a^2}{y^2} = \left( {{a^2} - {x^2}} \right)\left( {{a^2} + {\lambda ^2}} \right)$
$\dfrac{{\left( {{a^2} - {x^2}} \right)}}{{{a^2}{y^2}}} = \dfrac{1}{{{a^2} + {\lambda ^2}}}$
$\dfrac{{{a^2}}}{{{a^2}{y^2}}} - \dfrac{{{x^2}}}{{{a^2}{y^2}}} = \dfrac{1}{{{a^2} + {\lambda ^2}}}$
Now we have constants on one side and variables on one. By differentiating, we can easily eliminate $\lambda $
$\dfrac{1}{{{y^2}}} - \dfrac{1}{{{a^2}}}\left( {\dfrac{{{x^2}}}{{{y^2}}}} \right) = \dfrac{1}{{{a^2} + {\lambda ^2}}}$
Differentiating the above equation w.r.t $x$
$\dfrac{d}{{dx}}\left( {\dfrac{1}{{{y^2}}}} \right) - \dfrac{d}{{dx}}\left\{ {\dfrac{1}{{{a^2}}}\left( {\dfrac{{{x^2}}}{{{y^2}}}} \right)} \right\} = \dfrac{d}{{dx}}\left( {\dfrac{1}{{{a^2} + {\lambda ^2}}}} \right)$
Now, we know that derivative of a constant is zero
$\dfrac{d}{{dx}}\left( {\dfrac{1}{{{a^2} + {\lambda ^2}}}} \right) = 0$ …(i)
Also, $\dfrac{d}{{dx}}\left( {\dfrac{1}{{{y^2}}}} \right) = \dfrac{d}{{dx}}\left( {{y^{ - 2}}} \right)$
Now, derivative of ${x^n}$ is $n{x^{n - 1}}$
$\dfrac{d}{{dx}}\left( {\dfrac{1}{{{y^2}}}} \right) = - 2{y^{ - 3}}\dfrac{{dy}}{{dx}}$
$\dfrac{d}{{dx}}\left( {\dfrac{1}{{{y^2}}}} \right) = - 2{y^{ - 3}}{y^1} = - \dfrac{{2{y_1}}}{{{y^3}}}$ …(ii)
Now, $\dfrac{d}{{dx}}\left\{ {\dfrac{1}{{{a^2}}}\left( {\dfrac{{{x^2}}}{{{y^2}}}} \right)} \right\} = \dfrac{1}{{{a^2}}}\dfrac{d}{{dx}}\left( {\dfrac{{{x^2}}}{{{y^2}}}} \right)$
Now, applying the division rule, i.e.
$\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\left( {u'} \right) - u\left( {v'} \right)}}{{{v^2}}}$ where, $u'$and $v'$is the derivative of u and v with respect to $x$
Using this rule to find, $\dfrac{d}{{dx}}\left( {\dfrac{{{x^2}}}{{{y^2}}}} \right)$
$\dfrac{d}{{dx}}\left( {\dfrac{{{x^2}}}{{{y^2}}}} \right) = \dfrac{{\left( {{y^2}} \right)\left( {\dfrac{d}{{dx}}\left( {{x^2}} \right)} \right) - \left( {{x^2}} \right)\left( {\dfrac{d}{{dx}}\left( {{y^2}} \right)} \right)}}{{{{\left( {{y^2}} \right)}^2}}}$
$\dfrac{d}{{dx}}\left( {\dfrac{{{x^2}}}{{{y^2}}}} \right) = \dfrac{{\left( {{y^2}} \right)2x - \left( {{x^2}} \right)2y\dfrac{{dy}}{{dx}}}}{{{y^4}}}$
$\dfrac{d}{{dx}}\left( {\dfrac{{{x^2}}}{{{y^2}}}} \right) = \dfrac{{2x}}{{{y^2}}} - \dfrac{{2{x^2}}}{{{y^3}}}{y_1}$
$\dfrac{1}{{{a^2}}}\dfrac{d}{{dx}}\left( {\dfrac{{{x^2}}}{{{y^2}}}} \right) = \dfrac{1}{{{a^2}}}\left( {\dfrac{{2x}}{{{y^2}}} - \dfrac{{2{x^2}}}{{{y^3}}}{y_1}} \right)$ …(iii)
Now, $\dfrac{d}{{dx}}\left( {\dfrac{1}{{{y^2}}}} \right) - \dfrac{d}{{dx}}\left\{ {\dfrac{1}{{{a^2}}}\left( {\dfrac{{{x^2}}}{{{y^2}}}} \right)} \right\} = \dfrac{d}{{dx}}\left( {\dfrac{1}{{{a^2} + {\lambda ^2}}}} \right)$
Substituting, (i), (ii) and (iii)
$ - \dfrac{{2{y_1}}}{{{y^3}}} - \dfrac{1}{{{a^2}}}\left( {\dfrac{{2x}}{{{y^2}}} - \dfrac{{2{x^2}}}{{{y^3}}}{y_1}} \right) = 0$
$\dfrac{{2{y_1}}}{{{y^3}}} = \dfrac{1}{{{a^2}}}\left( {\dfrac{{2{x^2}}}{{{y^3}}}{y_1} - \dfrac{{2x}}{{{y^2}}}} \right)$
$\dfrac{{2{a^2}{y_1}}}{{{y^3}}} = \dfrac{{2{x^2}{y_1} - 2xy}}{{{y^3}}}$
Now, as we know$y \ne 0$
$2{a^2}{y_1} = 2{x^2}{y_1} - 2xy$
\[2xy = 2{x^2}{y_1} - 2{a^2}{y_1}\]
\[xy = \left( {{x^2} - {a^2}} \right){y_1}\]
$({x^2} - {a^2}){y_1} = xy$
Hence, the differential equation of given family of curves $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{a^2} + {\lambda ^2}}} = 1$is $({x^2} - {a^2}){y_1} = xy$
Therefore, option A $({x^2} - {a^2}){y_1} = xy$ is correct
Note: A different approach can be by making an estimation of the differential equation by looking at the number of arbitrary constants present in the given family of curves. The number of arbitrary constants present indicates the degree of the differential equation.
In the given family of curves $\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{a^2} + {\lambda ^2}}} = 1$, there is only a single arbitrary constant. So the degree of the differential equation would be 1, therefore making options b and c as incorrect. As mentioned, the method does only estimation, not a perfect way to get the correct answer.